Linear Algebra and its Applications 420 (2007) 400–406www.elsevier.com/locate/laa

Lower bounds of the Laplacian spectrum of graphsbased on diameter

Mei Lu a,∗,1, Lian-zhu Zhang b,1, Feng Tian c,2

a Department of Mathematical Sciences, Tsinghua University, Beijing 100084, Chinab School of Mathematical Science, Xiamen University, Xiamen, Fujian 361005, China

c Institute of Systems Science, Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences,Beijing 100080, China

Received 17 April 2006; accepted 31 July 2006Available online 28 September 2006

Submitted by R.A. Brualdi

Abstract

Let G be a connected graph of order n. The diameter of G is the maximum distance between any twovertices of G. In the paper, we will give some lower bounds for the algebraic connectivity and the Laplacianspectral radius of G in terms of the diameter of G.© 2006 Elsevier Inc. All rights reserved.

AMS classification: 05C50; 15A18

Keywords: Algebraic connectivity; Laplacian spectral radius; Diameter

1. Introduction

Let G = (V , E) be a simple undirected graph with vertex set V (G) = {v1, v2, . . . , vn}. If G

is a path, then G is denoted Pn. For vi ∈ V , the degree of vi , written as d(vi), is the number ofedges incident with vi . For two vertices vi and vj (i /= j), the distance between vi and vj is the

∗ Corresponding author.E-mail addresses: mlu@math.tsinghua.edu.cn (M. Liu), lz_zhang@126.com (L.-z. Zhang), ftian@mail.iss.ac.cn

(F. Tian).1 Partially supported by NSFC (No. 10571105).2 Partially supported by NSFC (No. 10431020).

0024-3795/$ - see front matter ( 2006 Elsevier Inc. All rights reserved.doi:10.1016/j.laa.2006.07.023

M. Lu et al. / Linear Algebra and its Applications 420 (2007) 400–406 401

number of edges in a shortest path joining vi and vj . The diameter of a graph is the maximumdistance between any two vertices of G.

Let A(G) be the adjacency matrix of G and D(G) = diag(d(v1), d(v2), . . . , d(vn)) be thediagonal matrix of vertex degrees. The Laplacian matrix of G is L(G) = D(G) − A(G). Clearly,L(G) is a real symmetric matrix. From this fact and Geršgorin’s Theorem, it follows that itseigenvalues are non-negative real numbers. The eigenvalues of an n × n matrix M are denoted byλ1(M), λ2(M), . . . , λn(M), while for a graph G, we will use λi(G) = λi to denote λi(L(G)), i =1, 2, . . . , n and assume that λ1(G) � λ2(G) � · · · � λn−1(G) � λn(G). It is well known thatλn(G) = 0 and the algebraic multiplicity of zero as an eigenvalue of L(G) is exactly the numberof connected components of G [10], i.e., the second smallest eigenvalue λn−1(G) > 0 if and onlyif G is connected. This led Fiedler [2] to define it as the algebraic connectivity of G, which hasa relation to the classical connectivity parameters of a graph G—the vertex connectivity and theedge connectivity. The eigenvalues, λ1(G) (also called the Laplacian spectral radius of G) andλn−1(G), have received a great deal of attention (see, for example [5,7–13]) and some resultsinvolved the diameter of a graph. For example, Mohar [12] showed that λn−1 � 4

nd, Alon and

Milman [1] showed d � 2[√

(2�/λn−1) log2 n]

and Chung [4] gave d �⌈

log(n−1)log(1/(1−λn−1)

⌉, where

n, d and � are the order, diameter and maximum degree of the graph G, respectively. In Section2, we also consider the diameter of G and give lower bounds of the Laplacian spectral radius andalgebraic connectivity of G involving the diameter.

2. Lower bounds for the Laplacian eigenvalues

Let G be a simple connected graph and L(G) = D(G) − A(G) be the Laplacian matrix ofG. It is well known that λn(G) = 0 with eigenvector e = (1, 1, . . . , 1)T and λn−1(G) > 0 by G

being connected. Since L(G) is symmetric, by the Rayleigh–Ritz Theorem (see for example [6]),we have

λn−1(G) = minx⊥e,x /=0

xTL(G)x

xTx(1)

and

λ1(G) = maxx /=0

xTL(G)x

xTx. (2)

Now, we will give a sharp lower bound for λn−1(G) by using some ideas of Fiedler [3].

Theorem 1. Let G be a connected simple graph of order n, size m and diameter d. Then

λn−1(G) � 2n

2 + n(n − 1)d − 2md. (3)

Equality holds if and only if G = P3 or G is a complete graph.

Proof. Let l2(V ) be a vector space and x = (x1, x2, . . . , xn) ∈ � be an eigenvector for λn−1(G),where � is the set of all non-constant vectors x ∈ l2(V ). Using the facts that

∑v∈V (G) xv = 0

(orthogonality to the eigenvector of λn) and that∑

uv∈E(G)(xu − xv)2 = λn−1(G)

∑v∈V (G) x2

v ,we can show easily that

λn−1(G) = 2n

∑uv∈E(G)(xu − xv)

2∑u∈V (G)

∑v∈V (G)(xu − xv)2

. (4)

402 M. Lu et al. / Linear Algebra and its Applications 420 (2007) 400–406

Since∑

v∈V (G) xv = 0 and G is connected, we have∑

uv∈E(G)(xu − xv)2 /= 0. Then, from (4),

we have

λn−1(G) = 2n

∑vivj ∈E(G),i<j (xvi

− xvj)2

∑ni=1

∑nj=1(xvi

− xvj)2

= 2n

∑vivj ∈E(G),i<j (xvi

− xvj)2

2∑

i<j (xvi− xvj

)2

= n

1 + g, (5)

where

g =∑

vivj /∈E(G),i<j (xvi− xvj

)2

∑vivj ∈E(G),i<j (xvi

− xvj)2

. (6)

Assume that u0 and v0 are the vertices such that (xu0 − xv0)2 = maxu,v∈V (G)(xu − xv)

2. Thenxu0 /= xv0 . Let P = v1v2 · · · vr+1 be the shortest path in G joining u0 and v0 (where v1 = u0 andvr+1 = v0). Note that∑

vivj ∈E(G),i<j

(xvi− xvj

)2 �∑

vivj ∈E(P ),i<j

(xvi− xvj

)2 (7)

�1

r(xv1 − xvr+1)

2 (8)

� 1

d(xu0 − xv0)

2

by using the Cauchy–Schwarz inequality. On the other hand, we have that∑

vivj /∈E(G),i<j

(xvi− xvj

)2 �(

n(n − 1)

2− m

)(xu0 − xv0)

2. (9)

Therefore, from (6) up to (9), we have

g �(

n(n − 1)

2− m

)d.

Thus, by (5), (3) holds.In order for the equality to hold, the inequalities from (7) up to (9) should be equalities. Then

r = d. By (7), we have that

xu − xv = 0 (10)

for any uv ∈ E(G) − E(P ). By (8), we have

xvi− xvi+1 = a /= 0 (11)

for all vivi+1 ∈ E(P ), where a is a constant (since (xv1 − xvd+1)2 = (xu0 − xv0)

2 /= 0, a /= 0 isobvious).

If the set {uv /∈ E(G)|u, v ∈ V (G)} = ∅, then G is a complete graph. Hence, we assume that{uv /∈ E(G)|u, v ∈ V (G)} /= ∅. Then r = d � 2. Note that v1vr+1 /∈ E(G) and xv1 /= xvr+1 , wehave g /= 0. From (9), for any uv �∈ E(G), we have

(xu − xv)2 = (xu0 − xv0)

2. (12)

M. Lu et al. / Linear Algebra and its Applications 420 (2007) 400–406 403

Now we show that d � 2. If d � 3, then we have v1vd, v1vd+1, vd−1vd+1 �∈ E(G) by P beingthe shortest path connecting v1 and vd+1. From (12), we have

(xv1 − xvd)2 = (xv1 − xvd+1)

2 = (xvd−1 − xvd+1)2. (13)

First we prove that xv1 /= xvd−1 . If d = 3, then from (11), we have xv1 /= xvd−1 . Thus we assumethat d � 4. Then v1vd−1 /∈ E(G) by P being the shortest path. From (12), we have

(xv1 − xvd)2 = (xv1 − xvd+1)

2 = (xv1 − xvd−1)2. (14)

If xv1 = xvd−1 , then (xv1 − xvd)2 = (xv1 − xvd+1)

2 = 0 by (14), i.e., xvd= xvd+1 , a contradiction

with (11). Hence xv1 /= xvd−1 .Now, by (13), we have

2xv1 = xvd+ xvd+1 and 2xvd+1 = xv1 + xvd−1 .

Therefore, we have 3(xv1 − xvd+1) = xvd− xvd−1 . Note that

(xvd− xvd−1)

2 � (xv1 − xvd+1)2 = (xu0 − xv0)

2 /= 0.

But we have

9(xv1 − xvd+1)2 = (xvd

− xvd−1)2 � (xv1 − xvd+1)

2

a contradiction. Hence, d � 2.Suppose that V (G) − V (P ) /= ∅. Thus, by the connectedness of G, there exists u ∈ V (G) −

V (P ) and some i, 1 � i � d + 1, such that uvi ∈ E(G). Then xu = xviby (10). If i = 1, then

uv2 /∈ E(G) (otherwise xu = xv2 by (10) which implies that xv1 = xv2 , a contradiction with (11))and uv3 /∈ E(G) when d = 2 (otherwise xu = xv3 which implies that xv1 = xv3 , a contradictionwith (xv1 − xv3)

2 = (xu0 − xv0)2 /= 0). Thus uv1Pvd+1 is the shortest path of length d + 1 join-

ing u and vd+1, a contradiction. Thus, we may assume that i /= 1, and i /= d + 1 by the same argu-ment. So, we have thatd = 2 anduv2 ∈ E(G). Thenxu = xv2 . By (10) and (11),uv1, uv3 /∈ E(G).But by (12), we have xu0 = xv0 , a contradiction. Thus V (G) − V (P ) = ∅, and hence G = P3.

Conversely, let G be a complete graph or G = P3. Then the equality holds by an elementarycalculation.

This completes the proof of Theorem 1. �

Remark 2. In [12], Mohar showed that

λn−1 � 4

nd. (15)

The bounds of (3) and (15) are incomparable. However we can give some graphs to show that thelower bound (3) is better than (15) in some cases. For example, it is easy to check that the lowerbound (3) is better than (15) when G is a complete graph or G = P3. In fact, if the size m of agraph G satisfies m � n(n−2)

4 + 1, then, from (3), we have

λn−1 � 4

nd + 4−4dn

� 4

nd.

Thus the lower bound (3) is better than (15). On the other hand, if m � n(n−2)4 , then from (15),

we have

λn−1 � 4n

n2d� 4n

2n(n − 1) − 4md>

2n

2 + n(n − 1)d − 2md.

Thus the lower bound (15) is better than (3).

404 M. Lu et al. / Linear Algebra and its Applications 420 (2007) 400–406

Next, we will give some lower bounds of λ1(G) involving the diameter. Let P be a path of G.We call P an induced path if the subgraph induced by V (P ) in G is P itself, i.e., G[V (P )] = P .Obviously, the shortest path between any two distinct vertices of G is an induced path.

Theorem 3. Let P = v1v2 · · · vs+1 be an induced path of G. Then

λ1(G) �∑s+1

i=1di + 2s

s + 1. (16)

Proof. Let the vertices of V (G) − V (P ) be labelled by vs+2, vs+3, . . . , vn if s < n − 1. Denotey = (y1, y2, . . . , yn)

T such that

yi =⎧⎨⎩

1 if i = 1, 3, . . . , s,

−1 if i = 2, 4, . . . , s + 1,

0 otherwise,

if s is an odd number, and

yi =⎧⎨⎩

1 if i = 1, 3, . . . , s + 1,

−1 if i = 2, 4, . . . , s,

0 otherwise,

if s is an even number. Obviously, y /= 0 and yTy = s + 1. Since P = v1v2 · · · vs+1 is an inducedpath, we have

yTL(G)y = (d(v1) + 1) +s∑

i=2

(d(vi) + 2) + (d(vs+1) + 1)

=s+1∑i=1

d(vi) + 2(s − 1) + 2

=s+1∑i=1

di + 2s.

Thus by (2), (16) holds. �

Remark 4. In [5], Grone and Merris had showed that

λ1 � � + 1, (17)

where � is the maximum degree of the graph G. We can easily find some graphs, for example,G is a path with at least 5 vertices or G is a regular graph, to show that the lower bound (16) isbetter than (17). In fact, if G exists an edge uv such that d(u) = d(v) = �, then the bound (17)can be derived by Theorem 3.

Let P be a path of G. Set d(P ) = ∑v∈P dG(v), where dG(v) is the degree of v. Denote by Pk

the set of the induced paths of length k. Then, by Theorem 3, we can easily show the followingresult.

Corollary 5. Let G be a connected graph of order n with diameter d. Then

λ1(G) � max1�k�d

maxP∈Pk

{d(P ) + 2k

k + 1

}.

M. Lu et al. / Linear Algebra and its Applications 420 (2007) 400–406 405

Let G be a graph of order n with vertices of degrees d1 � d2 � · · · � dn. Set

er = 1

r(d1 + d2 + · · · + dr), 1 � r � n.

Then by Corollary 5, the following result holds immediately.

Corollary 6. Let G be a connected graph of order n with diameter d. Then

λ1(G) � (d + 1)ed+1 + 2d

d + 1. (18)

Remark 7. The inequality (18) is sharp. Equality holds, for example, G is a complete graph.Since d(vi) � 2 for i = 2, . . . , d and d(v1), d(vd+1) � 1, we have the following result by

Corollary 6.

Corollary 8. Let G be a connected graph of order n with diameter d. Then

λ1(G) � 4d

d + 1.

Note that er is a monotonously non-decreasing function of r, i.e., er � er+1, we have the followingresult by Corollary 6.

Corollary 9. Let G be a connected graph of order n with diameter d, minimum degree δ(G).

Then

λ1(G) � (d + 1)δ(G) + 2d

d + 1.

Particularly, when G is k-regular, we have

λ1(G) � (d + 1)k + 2d

d + 1.

Acknowledgment

Many thanks to the anonymous referee for his/her many helpful comments and suggestions,which have considerably improved the presentation of the paper.

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